Astronuc said:Is this what one intended to write, or is this given in some text?
[tex]L= \frac{1}{2} (\partial_{\mu} \Psi)(\partial^{\mu} \Psi) + e^{-{(a \Psi)}^2} [/tex]
Lecticia said:Yes, exactly this Lagrangian, where \Psi is a scalar field.
dextercioby said:Well, just one question, you have the lagrangian, what are the field eqn's ?
nrqed said:This is a nonlinear field theory so, strictly speaking, there is no clear meaning for a mass term.
But I am guessing that they want you to treat the parameter "a" as small and to do a Taylor expansion of the exponential. If you do that, you will generate a mass term.
That's my guess.
The presence of a mass term in a field can be determined by analyzing the Lagrangian of the field. A mass term appears as a quadratic term in the Lagrangian, with a coefficient representing the mass of the field.
A mass term in a field indicates that the field has a non-zero rest mass, which means it cannot travel at the speed of light. This has important implications for the behavior of the field and its interactions with other particles.
No, a field with a mass term is by definition a massive field and cannot be considered massless. A massless field does not have a mass term in its Lagrangian and travels at the speed of light.
The presence of a mass term can break the symmetry of a field. For example, if a field has a mass term and also has a symmetry under rotations, the mass term will break this symmetry and the field will no longer behave the same way under rotations.
Yes, it is possible to add a mass term to a field without breaking any symmetries. This is known as a spontaneous symmetry breaking, where the field itself remains symmetric, but the vacuum state of the field is not.